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I am trying to do the problem: is $\mathbb{F}_{2011^2}[x] /(x^4 -6x -12)$ a field?

I know that this is a field if and only if $(x^4 - 6x - 12)$ is a maximal ideal, if and only if $x^4 - 6x - 12$ is irreducible over $\mathbb{F}_{2011^2}$. I also tried to use Rabin's algorithm:

Let $p_1, \ldots, p_k$ be all the prime divisors of $n$, and denote $n / p_i = n_i$, for $1 \leq i \leq k$. A polynomial $f \in \mathbb{F}_q [x]$ of degree $n$ is irreducible in $\mathbb{F}_q [x]$ if and only if gcd $(f, x^{q^ni} - x \mod f )=1$, for $1 \leq i \leq k$, and $f$ divides $x^{q^n}-x$.

In this case that would mean checking if $x^4 - 6x - 12$ divides $x^{2011^8}-x$ which doesn't sound like the most efficient thing to do.

Any help is much appreciated!

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1 Answer 1

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It can't be irreducible. The main obstacle is that it has coefficients in the very bottom field $\mathbb F_{2011}$, as well as the fact that a finite field has one and only one extension of each degree (inside a fixed algebraic closure).

First, if it is reducible over $\mathbb F_{2011}$ then it is a fortiori reducible over $\mathbb F_{2011^2}$. But if it is irreducible over $\mathbb F_{2011}$ then a root of it generates a degree $4$ extension of $\mathbb F_{2011}$. This degree $4$ extension must contain the unique degree $2$ extension, $\mathbb F_{2011^2}$, so it is a degree 2 extension of that field. Thus $f$'s roots have degree at most $2$ over the latter field, and so $f$ cannot be irreducible, since irreducibility would tell you that its roots degree $4$.

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    $\begingroup$ Indeed. Perhaps it deserves emphasis for beginners that we need to use the uniqueness of extensions of finite fields of given degrees... quite unlike number fields and such. $\endgroup$ Jul 28, 2020 at 22:09
  • $\begingroup$ @paulgarrett good point $\endgroup$
    – user208649
    Jul 28, 2020 at 22:15
  • $\begingroup$ I see! Thanks for the explanation! $\endgroup$
    – Sonja
    Jul 28, 2020 at 22:43

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